Question

Standard basis vectors for R^3 are (1,0,0),(0,1,0) and (0,0,1). If we want to insert u → into this basic, then which vector from standard basis can be removed while still maintaining the basis of R^3.
Discuss the case when:
u → =(4,3,6)
u → =(4,0,6)
Interpret the result geometrically in both cases.

Answer 1

Answer on Question #44340 – Math - Linear Algebra

Problem

Standard basis vectors for $R^3$ are $(1,0,0),(0,1,0)$ and $(0,0,1)$. If we want to insert $u \to$ into this basic, then which vector from standard basis can be removed while still maintaining the basis of $R^3$.

Discuss the case when:

$$u \rightarrow = (4, 3, 6)$$ $$u \rightarrow = (4, 0, 6)$$

Interpret the result geometrically in both cases.

Solution

1) $\vec{u} = (4,3,6)$.

$\vec{u} = 4(1,0,0) + 3(0,1,0) + 6(0,0,1)$, thus we can remove each vector from standard basis and each of $(\vec{u},\vec{e_1},\vec{e_2}),(\vec{u},\vec{e_1},\vec{e_3}),(\vec{u},\vec{e_3},\vec{e_2})$ will form the basis of $\mathbb{R}^3$.

Geometrical interpretation: none of $Oxy, Oyz, Oxz$ contains $\vec{u}$.

2) $\vec{u} = (4,0,6)$.

$\vec{u} = 4(1,0,0) + 6(0,0,1)$, which means that $(\vec{u},\vec{e_1},\vec{e_3})$ are not linearly independent, thus we can remove only $\vec{e_1}$ or $\vec{e_3}$ not $\vec{e_2}$. So, $(\vec{u},\vec{e_1},\vec{e_2}),(\vec{u},\vec{e_3},\vec{e_2})$ can form the basis of $\mathbb{R}^3$.

Geometrical interpretation: neither $Oxy$ nor $Oyz$ contains $\vec{u}$, but $\vec{u}$ lies in $Oxz$.

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